Answer:
dimensions of k are kg / s^2
Explanation:
F=ma (Newton's second law)
F=-kx (Hooke's law)
so equate both
-kx = ma
k = -ma/x
= mass * acceleration / distance (dimensions, -1 has no dimension)
= kg * m/s^2 / m
= kg / s^2
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
A stretched string is 120 cm long and has a linear density of 0.022 g/cm. What tension in the string will result in a second harmonic of 590 Hz
Answer:
T = 15,576 N
Explanation:
The speed of a wave on a string is given by
v = √ T /ρ rho
also the speed of the wave is given by the relationship
v = λ f
we substitute
λ f = √ T /ρ
T = (lam f)² ρ
let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is
L= λ/2 n
λ= 2L / n
we substitute
T = (2L / n f)²ρ rho
let's calculate
T = (2 1.20 / 2 590) 0.022
T = 15,576 N
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3800.0 μF and is charged to a potential difference of 78.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.84 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 4 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?
Answer:
- E = 11.55J
- Q = 0.17C
- E' = (1/4)E
Explanation:
- To calculate the amount of energy stored in the capacitor, you use the following formula:
[tex]E=\frac{1}{2}CV^2[/tex]
C: capacitance = 3800.0*10^-6F
V: potential difference = 78.0V
[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]
The energy stored in the capacitor is 11.55J
- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:
[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]
The charge on the capacitor is 0.17C
- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:
[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]
A: area of the plates
d: distance between plates
If the distance between plates is increased by a factor of 4, you have:
[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]
Then, the stored energy in the capacitor is decreased by a a factor of (1/4)
A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences
Answer:
The force experienced is 0.6 N
Explanation:
Given data
length of wire L= 2 m
current in wire I= 0.6 A
magnetic field B= 0.5
The force experienced can be represented as
[tex]F= BIL[/tex]
[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
Answer:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Explanation:
The force of gravity between two planets, is given by the following formula:
F = Gm₁m₂/r² ----------- equation 1
where,
F = Force of gravity between two planets
G = Gravitational Constant
m₁ = Mass of one planet
m₂ = Mass of other plant
r = Distance between two planets
Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:
F' = Gm₁m₂/(4r)²
F' = (1/16) (Gm₁m₂/r²)
using equation 1:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
9. A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is _____, if the ruler is assumed to have uniform mass.
Answer:
3.85 percent
Explanation:
From the question,
Percentage error = (error/actual)×100................ Equation 1
Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm
Substitute these values into equation 1
Percentage error = (0.6/15.6)×100
Percentage error = 3.85 percent
Hence the percentage error of the uniform mass = 3.85 percent
Two eggs of equal mass are thrown at a blanket with equal velocity. Egg B hits the blanket but egg A hits the wall instead. Compare the work done on the eggs in reducing their velocities to zero.
1. More work was done on A than on B.
2. It is meaningless to compare the amount of work because the forces were so different.
3. Work was done on B, but no work was done on A because the wall did not move.
4. More work was done on B than on A.
5. The amount of work is the same for both.
Answer:
5. The amount of work is the same for both.
Explanation:
Work done is a measure of change in kinetic energy of each egg
For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy
For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.
So work done is same for both eggs since they have same change in kinetic energy.
Saturn has a diameter of 74,900 miles. As the planet rotates on its axis, a point on its equator travels at 527,787 miles per day.
a. Find the angular speed of a point on its equator in radians per day.
b. Find the number of rotations the planet makes per day.
Answer:
a. ω = 14.1 radians per day
b. ω = 2.24 rotations per day
Explanation:
a.
The relationship between linear and angular velocity is given as:
v = rω
where,
v = linear speed of point = 527787 miles/day
r = radius of Saturn = diameter/2 = 74900 miles/2 = 37450 miles
ω = angular velocity of point = ?
Therefore,
527787 miles/day = (37450 miles)ω
ω = (527787 miles/day)/(37450 miles)
ω = 14.1 radians per day
b.
The number of rotations per day can be found by converting the units of angular acceleration:
ω = (14.1 radians per day)(1 rotation/2π radians)
ω = 2.24 rotations per day
Defination coulomb's law.
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256m/s
Explanation:
[tex]1.2 \mathrm{N} ; 2 \mathrm{N}[/tex]
2.[tex]200 \mathrm{N} ; 200 \mathrm{N}[/tex]
4.[tex]2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}[/tex]
[tex]5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}[/tex]
[tex]6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}[/tex]
[tex]8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}[/tex]
In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of [tex]256 \mathrm{m} / \mathrm{s}[/tex]
You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).
In both cases, the net force applied to the object must equal zero.
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!! In California, Clay is surfing on a wave that propels him toward the beach with a speed of 5.0 m/s. The wave crests are each 20 m apart. What is the frequency of the water wave? (please show your work and equation used)
Answer:
Frequency= 0.25m
Period= 4.0 secs
Explanation:
Clay is surfing on a wave with a speed of 5.0m/s
The wave crests are 20m apart
Therefore, the frequency of the wave can be calculated as follows
Frequency= wave speed/distance
= 5.0/20
= 0.25m
The period (T) can be calculated as follows
T= 1/frequency
T = 1/0.25
T= 4.0secs
Hence the frequency is 0.25m and the period is 4.0 secs
Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec
Answer:
3g/(8π²) ≈ 0.372 m
Explanation:
Draw a free body diagram. There is a weight force at the center of the pendulum.
Sum of the torques about the pivot:
∑τ = Iα
mg (½ L sin θ) = (⅓ mL²) α
3g sin θ = 2L α
α = 3g/(2L) sin θ
For small θ, sin θ ≈ θ.
α = 3g/(2L) θ
θ" = 3g/(2L) θ
The solution of this differential equation is:
θ = θ₀ cos(√(3g/(2L)) t)
So the period is:
T = 2π / √(3g/(2L))
If the period is 1 second:
1 = 2π / √(3g/(2L))
√(3g/(2L)) = 2π
3g/(2L) = 4π²
L = 3g/(8π²)
L ≈ 0.372 m
The length of the pendulum rod is 0.37 m.
What is meant by time period of a pendulum ?The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.
Here,
The mass of the pendulum, m = 2 kg
Time period of the pendulum, T = 1 s
Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.
The length of pendulum at that point = L/2
The perpendicular distance at that point, r = (L/2) sinθ
Therefore, the torque acting on the pendulum at that point,
τ = Iα
where I is the moment of inertia of the pendulum and α is the angular acceleration.
mg (L/2 sinθ) = (mL²/3)α
1/2 gsinθ = 1/3 Lα
Therefore,
α = (3g/2L) sinθ
For smaller values of θ, we can take sinθ = θ
So, α = (3g/2L) θ
We know that α = θ''
where θ is the angular displacement.
Therefore,
θ'' = (3g/2L) θ
So, ω = √3g/2L
Therefore, the equation of motion of the pendulum can be written as,
θ = θ₀ cos(ωt)
θ = θ₀ cos [(√3g/2L) t]
So, time period of the pendulum,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex]/(√3g/2L)
2[tex]\pi[/tex]/(√3g/2L) = 1
(√3g/2L) = 2[tex]\pi[/tex]
Therefore, length of the rod,
L = 3g/8[tex]\pi[/tex]²
L = 0.37 m
Hence,
The length of the pendulum rod is 0.37 m.
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Attaching the image file here.
59-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when: the elevator is descending at a constant rate of 8 m/s
Answer:
578.2 N
Explanation:
The computation of reading on the scale is shown below:-
Data provided in the question
Weight of a girl = 59 kg
Constant rate = 8 m/s
Since the elevator is descended so the acceleration is zero
As we know that
Reading on the scale is
[tex]F = m\times g[/tex]
where, m = 59 kg
g [tex]= 59 \times 9.8 m/s^2[/tex]
So, the reading on the scale is
= 578.2 N
Therefore for computing the reading on the scale we simply applied the above formula.
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Answer:
(a) 62.3 N
(b) 1.89 N
(c) 0.430 kg m²
Explanation:
(a) Find the acceleration of block B.
Δy = v₀ t + ½ at²
1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²
a = 0.90 m/s²
Draw a free body diagram of block B. There are two forces:
Weight force mg pulling down,
and tension force Tb pulling up.
Sum of forces in the -y direction:
∑F = ma
mg − Tb = ma
Tb = m (g − a)
Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)
Tb = 62.3 N
(b) Draw a free body diagram of block A. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and tension force Ta pulling right.
Sum of forces in the +x direction:
∑F = ma
Ta = ma
Ta = (2.10 kg) (0.90 m/s²)
Ta = 1.89 N
(c) Draw a free body diagram of the pulley. There are two forces:
Tension force Tb pulling down,
and tension force Ta pulling left.
Sum of torques in the clockwise direction:
∑τ = Iα
Tb r − Ta r = Iα
(Tb − Ta) r = I (a/r)
I = (Tb − Ta) r² / a
I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)
I = 0.430 kg m²
The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].
Given :
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.a) First, determine the acceleration of the B block.
[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]
[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]
[tex]\rm a = 0.9\; m/sec^2[/tex]
Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.
[tex]\rm \sum F=ma[/tex]
[tex]\rm mg-T_b=ma[/tex]
[tex]\rm T_b = m(g-a)[/tex]
[tex]\rm T_b = 7\times (9.8-0.9)[/tex]
[tex]\rm T_b = 62.3\;N[/tex]
b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.
[tex]\rm \sum F=ma[/tex]
[tex]\rm T_a=ma[/tex]
[tex]\rm T_a = 2.1\times 0.9[/tex]
[tex]\rm T_a = 1.89\;N[/tex]
c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.
[tex]\rm \sum \tau = I\alpha[/tex]
[tex]\rm T_br-T_ar = I\alpha[/tex]
[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]
[tex]\rm I = 0.430 \; kg\;m^2[/tex]
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What length (in m) should an oboe have to produce a fundamental frequency of 294 Hz on a day when the speed of sound is 350 m/s? It is open at both ends.
Answer:
L = 0.60 m
The length in metres should be 0.60 m
Explanation:
A pipe open at both ends can have a standing wave pattern with resonant frequency;
f = nv/2L ........1
Where;
v = velocity of sound
L = length of pipe
n = 1 for the fundamental frequency f1
Given;
Fundamental frequency f1 = 294 Hz
Velocity v = 350 m/s
n = 1
From equation 1;
Making L the subject of formula;
L = nv/2f1
Substituting the given values;
L = 1×350/(2×294)
L = 0.595238095238 m
L = 0.60 m
The length in metres should be 0.60 m
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
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Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3 g to cause it to rise into the air with an upward acceleration of 0.19 m/s2?
Answer:
q = 2.997*10^-4C
Explanation:
In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.
You use the second Newton law to sum both electrical and gravitational forces:
[tex]F_e-W=ma\\\\qE-mg=ma[/tex] (1)
Fe: electric force
W: weight of the penny
q: required charge = ?
m: mass of the penny = 3g = 0.003kg
E: magnitude of the electric field = 100N/C
g: gravitational acceleration = 9.8m/s^2
a: acceleration of the penny = 0.19m/s^2
You solve the equation (1) for q, and replace the values of the other parameters:
[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]
It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to move this charge 50 cm along a path making an angle of 40
Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J
What tension must a 47.0 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 2.55 m/s?
Answer:
I hope it will help you....
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.
Answer:
V = 0.45 Volts
Explanation:
First we need to find the total current passing through the wire. That can be given by:
Total Current = I = (Current Density)(Surface Area of Wire)
I = (Current Density)(2πrL)
where,
r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m
L = Length of Wire = 6.5 m
Therefore,
I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]
I = 1.25 x 10⁻⁴ A
Now, we need to find resistance of wire:
R = ρL/A
where,
ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm
A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²
Therefore,
R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)
R = 0.36 Ω
From Ohm's Law:
Voltage = V = IR
V = (1.25 x 10⁻⁴ A)(0.36 Ω)
V = 0.45 Volts
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
g beats can be heard when Question 5 options: when sound waves interfere constructively when sound waves of same frequency interfere when sound waves of slightly different frequencies interfere when sound waves of different amplitudes interfere
Answer:
Explanation:
Beats are produced when two sound waves from two different sources having slightly different frequencies interfere . Due to both constructive and destructive interference , high and low intensity sound is heard at some regular interval . This is called rate of beat formation . Amplitudes of sound may be same or different . High intensity sound heard per unit time is called beat . It is equal to difference of frequencies of the sound waves interfering
each other .
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?
Answer:
6.44 s
Explanation:
Given:
v₀ = 119 m/s
v = 233 m/s
a = 17.7 m/s²
Find: t
v = at + v₀
(233 m/s) = (17.7 m/s²) t + (119 m/s)
t = 6.44 s
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm
Answer:
[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]
Explanation:
Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:
[tex]a_{r} = \omega^{2}\cdot R[/tex]
Where:
[tex]\omega[/tex] - Angular speed, measured in radians per second.
[tex]R[/tex] - Radius of rotation, measured in meters.
The angular speed is first determined:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where [tex]\dot n[/tex] is the angular speed, measured in revolutions per minute.
If [tex]\dot n = 3000\,rpm[/tex], the angular speed measured in radians per second is:
[tex]\omega = \frac{\pi}{30}\cdot (3000\,rpm)[/tex]
[tex]\omega \approx 314.159\,\frac{rad}{s}[/tex]
Now, if [tex]\omega = 314.159\,\frac{rad}{s}[/tex] and [tex]R = 0.1\,m[/tex], the resultant acceleration is then:
[tex]a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)[/tex]
[tex]a_{r} = 9869.588\,\frac{m}{s^{2}}[/tex]
If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:
[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]
A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule with an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.
Answer:
The force is [tex]F = 8*10^{-12} \ N[/tex]
Explanation:
From the question we are told that
The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]
The energy provided by a single ATP is [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]
The velocity of the kinesin is [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]
The power provided by the ATP in one second is mathematically represented as
[tex]P = E_{ATP} * R[/tex]
substituting values
[tex]P = 80 * 0.8*10^{-19 }[/tex]
[tex]P = 6.4 *10^{-18}J/s[/tex]
Now this power is mathematically represented as
[tex]P = F * v[/tex]
Where F is the force the kinesin is exerting
Thus
[tex]F = \frac{P}{v}[/tex]
substituting values
[tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]
[tex]F = 8*10^{-12} \ N[/tex]
The force exerted by the kinesin is 8 × 10-12 N.
Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.
Since;
Energy = 0.8 × 10-19 J/molecule
Number ATP molecules transported per second = 80 ATP molecules/s
Power = 0.8 × 10-19 J/molecule × 80 ATP molecules/s
Power = 6.4 × 10-18 J
Again, we know that;
Power = Force × Velocity
Velocity of the ATP molecules = 800 nm/s or 8 × 10-7 m/s
Force = Power/velocity
Force = 6.4 × 10-18 J/ 8 × 10-7 m/s
Force = 8 × 10-12 N
Learn more: https://brainly.com/question/11897796
Which statement about the pressure of fluids is true?
O A decrease in the volume of a container will decrease the pressure.
O A decrease in the number of collisions will increase the pressure.
O An increase in the temperature will increase the pressure.
O A decrease in the number of particles in the container will increase the pressure
Next
Answer:
O An increase in the temperature will increase the pressure.
Explanation:
Decrease in volume, more collisions, increase in temperature, more particles: all of these increase the pressure.
Answer: O An increase in the temperature will increase the pressure.
A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon's radius increasing when the radius is 7ft
Answer:
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Explanation:
Given;
Rate of inflation dV = 140pift^3/min
Radius r = 7 ft
Change in radius = dr
Volume of a spherical balloon is;
V = (4/3)πr^3
The change in volume can be derived by differentiating both sides;
dV = (4πr^2)dr
Making dr the subject of formula;
dr = dV/(4πr^2)
Substituting the given values;
dr = 140π/(4π×7^2)
dr = 0.714285714285 ft/min
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.